NEET-XII-Physics
11: Gravitation
- #27At what rate should the earth rotate so that the apparent g at the equator becomes zero? What will be the length of the day in this situation?Ans : The apparent acceleration due to gravity at the equator becomes zero.
i.e., g' = g - ω2R = 0
⇒ g = ω2R
`` \Rightarrow \,\mathrm{\,\omega \,}=\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6400\times {10}^{3}}}``
`` \Rightarrow \,\mathrm{\,\omega \,}=\sqrt{\frac{9.8\times {10}^{-5}}{6.4}}=\sqrt{1.5\times {10}^{-6}}``
`` \Rightarrow \,\mathrm{\,\omega \,}=1.2\times {10}^{-3}\,\mathrm{\,rad\,}/\,\mathrm{\,s\,}``
`` \therefore T=\frac{2\,\mathrm{\,\pi \,}}{\,\mathrm{\,\omega \,}}=\frac{2\times 3.14}{1.2\times {10}^{-3}}``
`` =\frac{6.28}{1.2\times {10}^{-3}}``
`` =1.41\,\mathrm{\,h\,}``
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